If P and $Q$ are the points of intersection of the circles

$x^2+y^2+3x+7y+2p-5=0\text{ and }{x}^2+y^2+2x+2y-p^2=0$

then there is a circle passing through$P, Q$and $(1, 1)$ for

The equation of the circle passing through$P$ and $Q$ is

$\left(x^2+y^2+3x+7y+2p-5\right)+\lambda \left(x^2+y^2+2x+2y-p^2\right)=0\dots \left(\mathrm{i}\right)$

$\left[\text{ Using: }{S}_1+\lambda S_2=0\right]$

It passes through $(1, 1).$

$\therefore (2p+7)+\lambda \left(6-p^2\right)=0\Rightarrow \lambda =\frac{2p+7}{p^2-6}$

if $p=-\frac{7}{2}$ then $\lambda =0$ and equation (i) represents first circle

If $p^2=6$ then equation (i) represents second circle. For all other values of $p$, equation represents some other circle. Hence, equation (i) represents a circle for all values of $p.$